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NCERT Solutions for Class 8 Maths Chapter 11: Mensuration - Exercise 11.4

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NCERT Solutions for Class 8 Maths Chapter 11 (EX 11.4)

Free PDF download of NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.4 (EX 11.4) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 8 Maths Chapter 11 Mensuration Exercise 11.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 8

Subject:

Class 8 Maths

Chapter Name:

Chapter 11 - Mensuration

Exercise:

Exercise - 11.4

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



Every NCERT Solution is provided to make the study simple and interesting. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 8 Science , Maths solutions and solutions of other subjects. You can also download NCERT Solutions for Class 8 Maths to help you to revise complete syllabus and score more marks in your examinations.

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Access NCERT Solutions for Class 8 Chapter 11- Mensuration

1. Given a cylindrical tank, in which situation will you find the surface area and in which situation volume.

a. To find how much it can hold.

Ans: To find the capacity of any object we find the volume. Here, see that it is asked how much the cylindrical tank can hold. Thus, we will find the volume of the tank in this case.

b. Number of cement bags required to plaster it

Ans: The walls of the tank will be cemented. This situation deals with the surface area. So, we will find the surface area of the cylinder to find the number of cement bags required to plaster it.

c. To find the number of smaller tanks that can be filled with water from it.

Ans: To find the capacity of any object we find the volume. Here, see that it is asked to find the number of smaller tanks that can be filled from the water. Thus, we will find the volume of the tank in this case.

2. The diameter of the cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without  doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has the greater area.

Ans: The volume of a cylinder with height $h$, and radius $r$ is $V = \pi {r^2}h$.

The diameter of cylinder A is 7 cm. 

The radius of cylinder A will be $\dfrac{7}{2}$ cm.

The diameter of cylinder B is 14 cm.

The radius of cylinder B will be 7 cm.

$ \Rightarrow \dfrac{7}{2} < 7$

For instance, if the value of $r$ and $h$ is the same, then the cylinder with the greater radius will have greater area.

We can see that the radius of cylinder B is greater, so, the volume of cylinder B will be greater. Now, verify it by calculating the volume of both the cylinders.

Assume the volume of cylinder A be ${V_A}$.

$\Rightarrow {V_A} = \pi  \times \dfrac{7}{2} \times \dfrac{7}{2} \times 14$

$ \Rightarrow {V_A} = \dfrac{{22}}{7} \times \dfrac{7}{2} \times \dfrac{7}{2} \times 14$

$ \Rightarrow {V_A} = 11 \times 7 \times 7$

$ \Rightarrow {V_A} = 539{\text{ c}}{{\text{m}}^3} $

Assume the volume of cylinder A be ${V_B}$.

$\Rightarrow {V_B} = \pi  \times 7 \times 7 \times 7 $

$ \Rightarrow {V_B} = \dfrac{{22}}{7} \times 7 \times 7 \times 7$

$ \Rightarrow {V_B} = 22 \times 49$

$ \Rightarrow {V_B} = 1078{\text{ c}}{{\text{m}}^3}  $

$ \Rightarrow {V_B} > {V_A}$

Hence, the volume of cylinder B is greater.

Now, we measure the surface area of both the cylinders to find which cylinder has a greater area. The formula of surface area of the cylinder be $L = 2\pi r\left( {r + h} \right)$.

Let the surface of the cylinder A be ${L_A}$.

$\Rightarrow {L_A} = 2 \times \dfrac{{22}}{7} \times \dfrac{7}{2}\left( {\dfrac{7}{2} + 14} \right)$

$\Rightarrow {L_A} = 2 \times 11 \times \left( {\dfrac{{7 + 28}}{2}} \right)$

$\Rightarrow {L_A} = 2 \times 11 \times \left( {\dfrac{{35}}{2}} \right)$

$\Rightarrow {L_A} = 11 \times 35$

$\Rightarrow {L_A} = 385\,{\text{c}}{{\text{m}}^2}$

Let the surface of cylinder A be ${L_B}$.

$\Rightarrow {L_B} = 2 \times \dfrac{{22}}{7} \times 7 \times \left( {7 + 7} \right)$

$\Rightarrow {L_B} = 2 \times 22 \times 14$

$\Rightarrow {L_B} = 44 \times 14$

$\Rightarrow {L_B} = 616\,{\text{c}}{{\text{m}}^2} $

From the above calculations, ${L_{\text{B}}} > {L_{\text{A}}}$.

Hence, cylinder B has greater surface area than the surface area of cylinder A.

3. Find the height of a cuboid whose base area is 180 square cm and volume is 900 cubic cm?

Ans: The base of the cuboid is rectangular. Thus, the area of the base is a product of length and breadth.  Thus, $A = 180\,{\text{c}}{{\text{m}}^2}$.

The volume of the cuboid is $V = l \times b \times h$. We are given that the volume is 900 cubic cm. 

$\Rightarrow 900 = 180 \times {\text{height}} $

$\Rightarrow \dfrac{{900}}{{180}} = {\text{height}} $

$\Rightarrow 5 = {\text{height}} $ 

Hence, the required height is 5 cm.

4. A cuboid is of dimension $60{\text{cm}} \times 54{\text{cm}} \times 30{\text{cm}}$. How many small cubes with side 6 cm can be placed in the given cuboid?

Ans: The volume of the cuboid is $V = l \times b \times h$. 

$\Rightarrow {V_{{\text{cuboid}}}} = 60{\text{cm}} \times 54{\text{cm}} \times 30{\text{cm}} $

$\Rightarrow {V_{{\text{cuboid}}}} = 97200{\text{c}}{{\text{m}}^3} $ 

The cube has a side length of 6 cm. The volume of the cube is given by $V = {\left( {{\text{side}}} \right)^3}$.

$ \Rightarrow {V_{{\text{cube}}}} = {\left( 6 \right)^3} $

$\Rightarrow {V_{{\text{cube}}}} = 216\,{\text{c}}{{\text{m}}^3} $ 

Let the required number of cubes be $n$.

$\Rightarrow n = \dfrac{{{\text{volume of the cuboid}}}}{{{\text{volume of the cube}}}} $

$ \Rightarrow n = \dfrac{{{\text{97200}}}}{{{\text{216}}}} $

$\Rightarrow n = {\text{450}} $ 

Hence, the required number of cubes are 450.

5. Find the height of the cylinder whose volume is $1.54\,{{\text{m}}^3}$ and the diameter of the base is 140 cm?

Ans: The diameter is 140 cm.

The radius is half of the diameter.

$ \Rightarrow r = \dfrac{{140}}{2}{\text{cm}} $

$ \Rightarrow r = 70{\text{cm}} $ 

Convert cm into m.

$\Rightarrow r = \dfrac{{70}}{{100}}{\text{m}} $

$ \Rightarrow r = \dfrac{7}{{10}}{\text{m}} $ 

Now, we will find the volume of the cylinder using the formula, $V = \pi {r^2}h$. Here, we are given that the volume is $1.54$.

$ \Rightarrow 1.54 = \dfrac{{22}}{7} \times \dfrac{7}{{10}} \times \dfrac{7}{{10}} \times h $

$\Rightarrow 1.54 = 22 \times \dfrac{1}{{10}} \times \dfrac{7}{{10}} \times h $ 

Isolate the variable to one side and the rest of the numbers to the other.

$\Rightarrow h = \dfrac{{1.54 \times 10 \times 10}}{{22 \times 7}} $

$\Rightarrow h = 1\,{\text{m}} $ 

Hence, the height of the cylinder is 1 m.

6. A milk tank is in the form of cylinder whose radius is $1.5$ m and length is 7 m. find the quantity of milk in litres that can be sorted in the tank?

Ans: The radius, $r$ of the cylinder is $1.5$m.

The length, $h$ of the cylinder is 7m.

The volume of the cylinder is given by $V = \pi {r^2}h$.

$\Rightarrow V = \pi  \times {\left( {1.5} \right)^2} \times \left( 7 \right) $

$\Rightarrow V = \dfrac{{22}}{7} \times 1.5 \times 1.5 \times 7 $

$  \Rightarrow V = 22 \times 2.25 $

$\Rightarrow V = 49.5\,{{\text{m}}^3} $ 

Now, we know that, $1\,{{\text{m}}^3} = 1000\,{\text{L}}$. So, convert the volume into litre.

$\Rightarrow 49.5\,{{\text{m}}^3} = 49.5 \times 1000\,{\text{L}} $

$\Rightarrow 49.5\,{{\text{m}}^3} = 49500\,{\text{L}} $ 

Therefore, 49500 litre of milk can be stored in the cylindrical tank.

7. If each edge of the cube is double,

i. How many times will its surface area increase?

Ans: Let the edge of the cube be $l$. Let the initial surface area be ${S_1}$.

The initial surface area will be ${S_1} = 6{l^2}$.

When the edge of the cube is doubled, $l$ becomes $2l$.

Let the new surface area be ${S_2}$.

$ \Rightarrow {S_2} = 6{\left( {2l} \right)^2} $

$ \Rightarrow {S_2} = 6\left( {4l} \right) $  

Thus, the surface area will increase 4 times when the edge of the cube is doubled.

ii. How many times will its volume increase?

Ans: Let the edge of the cube be $l$. Let the initial volume be ${V_1}$.

The initial volume will be ${V_1} = {l^3}$.

When the edge of the cube is doubled, $l$ becomes $2l$.

Let the new surface area be ${V_2}$.

$\Rightarrow {V_2} = {\left( {2l} \right)^3} $

$\Rightarrow {V_2} = 8{l^3} $ 

Hence, the volume becomes 8 times of the original volume when the radius is doubled.

8. Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of the reservoir is 108 cubic m, find the number of hours it will take to fill the reservoir?

Ans: The volume of the reservoir is 108 ${{\text{m}}^{\text{3}}}$.

First, we will convert the volume into litres. We know that, $1\,{{\text{m}}^3} = 1000\,{\text{L}}$. 

$ \Rightarrow 108{{\text{m}}^3} = 108 \times 1000\,{\text{L}} $

$\Rightarrow 108{{\text{m}}^3} = 108000\,\,{\text{L}} $ 

The water is getting poured at the rate of 60 litre per minute.

The amount of water poured in 1 hour will be,

$ \Rightarrow 60 \times 60\,{\text{L = 3600}}\,{\text{L per hour}}$

Let $n$ be the number of hours.

We can get the number of hours by dividing the total volume of the reservoir by the amount of water being poured.

$\Rightarrow n = \dfrac{{108000}}{{3600}} $

$\Rightarrow n = 30\,{\text{hours}} $ 

Hence, 30 hours will be required to fill the reservoir.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (Ex 11.4) Exercise 11.4

Opting for the NCERT solutions for Ex 11.4 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.4 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 11 Exercise 11.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 11 Exercise 11.4, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 8 Maths Chapter 11 Exercise 11.4 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

FAQs on NCERT Solutions for Class 8 Maths Chapter 11: Mensuration - Exercise 11.4

1. Do i need to Practice all Questions Provided in NCERT Solutions Class 8 Maths Mensuration?

There are many essential mensuration Class 8 questions from Chapter 11. These maths problems include short answers, long answers, and HOTS questions, all of which are important for CBSE examinations. The questions in Class 8 for Chapter 11 are divided into two categories: quick answers and lengthy answers. These problems span over a variety of topics and will assist students in honing their problem-solving abilities in preparation for the test.

2. What are the Important Topics Covered in NCERT Solutions Class 8 Maths Chapter 11?

These NCERT Solutions for Class 8 Maths Chapter 11 available on the official website of Vedantu includes all of the essential mensuration formulae and properties of different geometric forms and figures and answer all of the problems relating to the principles of mensuration. Subject specialists address difficulties with the perimeter and area of other planar closed figures such as quadrilaterals. The PDFs of NCERT Solutions are also available on the Vedantu app at free of cost.

3. What is Mensuration Class 8 Maths?

Mensuration is the field of mathematics concerned with geometric forms and their characteristics such as length, volume, shape, surface area, lateral surface area, and so on. It is a branch that examines the computation of geometric figures and their characteristics such as area, length, volume, lateral surface area, and surface area, among others. It explains all of the basic equations and characteristics of numerous geometric forms and figures as well as the foundations of computation.

4. What are the shapes dealt with in Mensuration?

There are many shapes dealt with in the Chapter 11 of Class 8 Maths:

  • Cylinder.

  • Circles.

  • Polygons.

  • Rectangles and Squares.

  • Trapezium, Parallelogram and Rhombus.

  • Area and Perimeter.

  • Cube and Cuboid.

5. What are the chapters in Class 8 Maths?

1

Rational Numbers

2

Exponents and Powers

3

Squares and Squares Roots, Cubes and Cube Roots

4

Linear Equations in one variable

5

Understanding Quadrilaterals and Constructions

6

Mensuration

7

Algebraic Expressions, Identities and Factorization

8

Ratio, proportion and percentage

9

Direct and Inverse Proportion

10

Representing 3D in 2D

11

Playing with Numbers

12

Data Handling